NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

# NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Edited By Ramraj Saini | Updated on Sep 26, 2023 07:50 PM IST

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Linear Equations in Two Variables Class 9 Questions And Answers are provided here. These are prepared by expert team at careers360 keeping in mind latest CBSE syllabus 2023. In this particular chapter, you will study linear equations in two variables of the type ax + by + c = 0 where a, b and c are the real numbers, and a and b both are not zero. Class 9 linear equations in two variables NCERT solutions are there to make your task easy while preparing for the exams.

This class 9 linear equations in two variables always comes with a good number of questions in competitive exams like the Indian National Olympiad (INO), National Talent Search Examination (NTSE). Linear equations in two variables class 9 NCERT solutions are designed in such a manner that a student can get maximum marks assigned to that particular question. are also available chapter wise which can be downloaded by clicking on the given link.

## Linear Equations in Two Variables Class 9 Solutions - Important Formulae

• (a + b)2 = a2 + 2ab + b2

• (a - b)2 = a2 - 2ab + b2

• (a + b)(a - b) = a2 - b2

• (x + a)(x + b) = x2 + (a + b)x + ab

• (x + a)(x - b) = x2 + (a - b)x - ab

• (x - a)(x + b) = x2 + (b - a)x - ab

• (x - a)(x - b) = x2 - (a + b)x + ab

• (a + b)3 = a3 + b3 + 3ab(a + b)

• (a - b)3 = a3 - b3 - 3ab(a - b)

• (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz

• (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz

• (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz

• (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz

• x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

• x2 + y2 = 1/2 [(x + y)2 + (x - y)2]

• (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

• x3 + y3 = (x + y)(x2 - xy + y2)

• x3 - y3 = (x - y)(x2 + xy + y2)

• x2 + y2 + z2 - xy - yz - zx = 1/2 [(x - y)2 + (y - z)2 + (z - x)2]

Free download NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables for CBSE Exam.

## Linear Equations in Two Variables Class 9 NCERT Solutions (Intext Questions and Exercise)

Linear equations in two variables class 9 solutions Exercise: 4.1

Let the cost of a notebook be Rs x and that of a pen be Rs y .

According to the given condition: The cost of a notebook is twice the cost of a pen.

Thus, $x=2y$

$\Rightarrow x-2y=0$

Given : $2x + 3y = 9.3\bar{5}$

$\Rightarrow 2x + 3y - 9.3\bar{5}=0$

Here , a=2, b=3 and c = $- 9.3\bar{5}$

Given:

$x - \frac{y}{5} - 10 = 0$

$\Rightarrow x - \frac{y}{5} - 10 = 0$

Here ,

a=1,

$b=\frac{-1}{5}$

c = -10

Given :

$-2x + 3y = 6$

$\Rightarrow -2x + 3y - 6=0$

Here , a= -2, b=3 and c = -6

Given : $x =3y$

$\Rightarrow x -3y=0$

Here , a= 1, b= -3 and c =0

Given : $2x = - 5y$

$\Rightarrow 2x + 5y=0$

Here , a=2, b= 5 and c =0

Given : $3x + 2 = 0$

$\Rightarrow 3x + 2 = 0$

Here , a= 3, b=0 and c =2

Given : $y -2 = 0$

$\Rightarrow 0.x+y -2 = 0$

Here , a=0, b= 1 and c = -2

Given : $5 = 2x$

$\Rightarrow 2x+0.y-5=0$

Here , a=2, b= 0 and c = -5

Class 9 maths chapter 4 question answer Exercise: 4.2

### (iii) infinitely many solutions

Given : $y = 3x + 5$

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

(i) $2x + y = 7$ (ii) $\pi x +y = 9$ (iii) $x = 4y$

(i) Given : $2x + y = 7$

Putting x=0, we have , $y=7-2\times 0=7$ means $(0,7)$ is a solution.

Putting x=1, we have , $y=7-2\times 1=5$ means $(1,5)$ is a solution.

Putting x=2, we have , $y=7-2\times 2=3$ means $(2,3)$ is a solution.

Putting x=3, we have , $y=7-2\times 3=1$ means $(3,1)$ is a solution.

The four solutions are : $(0,7),(1,5),(2,3),(3,1)$ .

(ii) Given : $\pi x +y = 9$

Putting x=0, we have , $y=9-\pi \times 0=9$ means $(0,9)$ is a solution.

Putting x=1, we have , $y=9-\pi \times 1=9-\pi$ means $(1,9-\pi )$ is a solution.

Putting x=2, we have , $y=9-\pi \times 2=9-2\pi$ means $(2,9-2\pi )$ is a solution.

Putting x=3, we have , $y=9-\pi \times 3=9-3\pi$ means $(3,9-3\pi )$ is a solution.

The four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$ .

(iii) Given : $x = 4y$

Putting x=0, we have , $y=\frac{0}{4}=0$ means $(0,0)$ is a solution.

Putting x=1, we have , $y=\frac{1}{4}$ means $(1,\frac{1}{4})$ is a solution.

Putting x=2, we have , $y=\frac{2}{4}=\frac{1}{2}$ means $(2,\frac{1}{2})$ is a solution.

Putting x=3, we have , $y=\frac{3}{4}$ means $(3,\frac{3}{4})$ is a solution.

The four solutions are : $(0,0)$ , $(1,\frac{1}{4})$ , $(2,\frac{1}{2})$ and $(3,\frac{3}{4})$ .

(i) Given : $x - 2y = 4$

Putting $(0,2)$ ,

we have , $x - 2y = 0-2(2)=-4\neq 4$

Therefore, $(0,2)$ is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (2,0),

we have , $x - 2y = 2-2(0)=2\neq 4$

Therefore, (2,0) is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (4,0),

we have , $x - 2y = 4-2(0)=4=4$

Therefore, (4,0) is a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting $(\sqrt2 , 4\sqrt2)$ ,

we have , $x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4$

Therefore, $(\sqrt2 , 4\sqrt2)$ is not a solution of $x - 2y = 4$ .

Given : $x - 2y = 4$

Putting (1,1) ,

we have , $x - 2y = 1-2(1)=-1\neq 4$

Therefore, (1,1) is not a solution of $x - 2y = 4$ .

Given : $2x + 3y = k$

Putting (2,1),

we have , $k=2x + 3y = 2(2)+3(1)=4+3=7$

Therefore, k=7 for $2x + 3y = k$ putting x=2 and y=1.

Class 9 maths chapter 4 question answer Exercise: 4.3

Given : $x + y = 4$

Putting x=0,we have $y=4-0=4$

Putting x=1,we have $y=4-1=3$

Thus, (0,4) and (1,3) are solutions of given equation.

Given : $x - y = 2$

Putting x=0,we have $y=0-2=-2$

Putting x=1,we have $y=1-2=-1$

Thus, (0,-2) and (1,-1) are solutions of given equation.

Given : $y = 3x$

Putting x=0,we have $y=3(0)=0$

Putting x=1,we have $y=3(1)=3$

Thus, (0,0) and (1,3) are solutions of given equation.

Given : $3 = 2x + y$

Putting x=0,we have $y=3-2.(0)=3$

Putting x=1,we have $y=3-2.(1)=1$

Thus, (0,3) and (1,1) are solutions of given equation.

The equations of two lines passing through (2, 14) are given by : $x+y=16\, \, and\, \, x-y=-12$

There are infinite lines passing through (2, 14) because infinite lines pass through a point.

Given : the point (3, 4) lies on the graph of the equation $3y = ax+ 7$

Put x=3 and y=4

$3y = ax+ 7$

$\Rightarrow 3(4) = a(3)+ 7$

$\Rightarrow 12 = 3a+ 7$

$\Rightarrow 12 -7 = 3a$

$\Rightarrow 5 = 3a$

$\Rightarrow a=\frac{5}{3}$

Given: The distance covered as x km and total fare is Rs. y.

Total fare =the fare for first km + the fare of the remaining distance

$\therefore y=8+5\times (x-1)$

$\Rightarrow y=8+5x-5$

$\Rightarrow y=3+5x$

For graph,

Putting x=0, we have $y=3+5(0)=3$

Putting x=1, we have $y=3+5(1)=8$

Putting x=2, we have $y=3+5(2)=13$

Hence,(0,3),(1,8) and (2,13) are solutions of equation.

The graph is as shown :

(i) $y = x$

(ii) $x + y = 0$

(iii) $y = 2x$

(iv) $2 + 3 y = 7x$

For the given figure :

Points on line are (-1,1) , (0,0 ) and (1,-1)

$x + y = 0$ satisfies all the above points.

Thus, $x + y = 0$ is the correct equation of the line.

(i) $y = x + 2$

(ii) $y = x - 2$

(iii) $y = -x + 2$

(iv) $x + 2y = 6$

For the given figure :

Points on line are (-1,3) , (0,2 ) and (2,0)

$y = -x + 2$ satisfies all the above points.

Thus, $y = -x + 2$ is the correct equation of the line.

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. $y\, \, \alpha \, \, x$

$\Rightarrow y=kx$

k=Constant force = 5

Then,

$\Rightarrow y=5x$

For graph,

Put x=0,we have $y=5(0)=0$

Put x=1,we have $y=5(1)=5$

Put x=2,we have $y=5(2)=10$

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 2 units then the work done is 10 units.

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. $y\, \, \alpha \, \, x$

$\Rightarrow y=kx$

k=Constant force = 5

Then, $\Rightarrow y=5x$

For graph,

Put x=0,we have $y=5(0)=0$

Put x=1,we have $y=5(1)=5$

Put x=2,we have $y=5(2)=10$

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 0 units then work done is 0 units.

Let the contribution of Yamini be x.

contribution of Yamini be y.

According to question,

$x+y=100$

For x=0 , we have $y=100-0=100$

For x=10 , we have $y=100-10=90$

For x=20 , we have $y=100-20=80$

Hence, (0,100) , (10,90) and (20,80)

Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.

Let celsius be on x-axis and Fahrenheit be on y-axis.

$F = \left(\frac{9}{5} \right )C + 32$

For graph,

Putting x=0, we get $y= \left(\frac{9}{5} \right )(0)+ 32=32$

Putting x=5, we get $y= \left(\frac{9}{5} \right )(5)+ 32=41$

Putting x=10, we get $y= \left(\frac{9}{5} \right )(10)+ 32=50$

Hence, points are (0,32) , (5,41) and (10,50).

If the temperature is 30°C, what is the temperature in Fahrenheit?

$F = \left(\frac{9}{5} \right )C + 32$

Put c=30,

$F = \left(\frac{9}{5} \right )30 + 32=54+32=86$

Thus, the temperature = 30°C, then the temperature is 86 in Fahrenheit.

If the temperature is 95°F, what is the temperature in Celsius?

$F = \left(\frac{9}{5} \right )C + 32$

Put F=95,

$95 = \left(\frac{9}{5} \right )C + 32$

$\Rightarrow \left(\frac{9}{5} \right )C=95 -32$

$\Rightarrow \left(\frac{9}{5} \right )C=63$

$\Rightarrow C=35$

If the temperature is 95°F, then 35 is the temperature in Celsius.

If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

If the temperature is 0°C,

$F = \left(\frac{9}{5} \right )C + 32$

$F = \left(\frac{9}{5} \right )(0) + 32$

$\Rightarrow F = 32$

if the temperature is 0°F,

$F = \left(\frac{9}{5} \right )C + 32$

$0= \left(\frac{9}{5} \right )C + 32$

$\Rightarrow \left(\frac{9}{5} \right )C=- 32$

$\Rightarrow C=- 17.8$

Thus, if the temperature is 0°C , then the temperature in Fahrenheit is 32 and if the temperature is 0°F, then the temperature in celsius is -17.8.

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

$F = \left(\frac{9}{5} \right )C + 32$

Let temperature be x in both Fahrenheit and celsius.

$x= \left(\frac{9}{5} \right )x + 32$

$\Rightarrow 5x= 9x + 160$

$\Rightarrow 5x-9x = 160$

$\Rightarrow -4x = 160$

$\Rightarrow x = -40 \degree$

Thus, 40 is the temperature which is numerically the same in both Fahrenheit and celsius.

Class 9 maths chapter 4 NCERT solutions Exercise: 4.4

in one variable

Equation $y = 3$ can be represented in one variable on the number line.

For 2 variable representations of $y = 3$ .

Equation : $0.x+y=3$

For graph,

x=0,we have y=3

x=1,we have y=3

x=2,we have y=3

Hence, (0,3) ,(1,3) and (2,3) are solutions of equation.

Equation $2x + 9 = 0$ can be represented in one variable on the number line.

$\Rightarrow x=\frac{-9}{2}$

The yellow mark represents x= - 4.5.

For 2 variable representation of $2x + 9 = 0$ .

Equation : $2x +0.y =-9$

For graph,

y=0,we have $x=\frac{-9}{2}$

y=1,we have $x=\frac{-9}{2}$

y=2,we have $x=\frac{-9}{2}$

Hence, $(\frac{-9}{2},0)$ , $(\frac{-9}{2},1)$ and $(\frac{-9}{2},2)$ are solutions of equation.

## Linear Equations In Two Variables Class 9 NCERT Solutions - Summary

The chapter 4 maths class 9 Textbook is called "Linear Equations in Two Variables," which is included in Unit 2 Algebra. In the board exams, Algebra unit comprises a total of 20 marks, consisting of 1 multiple choice question for 1 mark, 2 short answers with reasoning for a total of 4 marks, 3 short answer questions for a total of 9 marks, and 1 long answer question for 6 marks. The chapter covers the following topics:

 Chapter 4 Linear Equations in Two Variables 4.1 Introduction 4.2 Linear Equations 4.3 Solution of a Linear Equation 4.4 Graph of a Linear Equation in Two Variables 4.5 Equations of Lines Parallel to the x-axis and y-axis

In this class 9 chapter 4 maths, there are a total of 4 exercises which consist of 16 questions. NCERT solutions for Class 9 Maths chapter 4 Linear Equations in Two Variables is covering the detailed solutions to each and every question present in the practice exercises. This is an important chapter as this created a foundation for the higher level of algebra. Algebra is a unit in class 9 maths which holds 20 marks in the final examination.

Interested students can practice class 9 maths ch 4 question answer using the links given below.

NCERT Solutions for Class 9 Maths - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations In Two Variables Chapter 5 Introduction to Euclid's Geometry Chapter 6 Lines And Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron’s Formula Chapter 13 Surface Area and Volumes Chapter 14 Statistics Chapter 15 Probability

## NCERT Solutions for Class 9 - Subject Wise

 NCERT Solutions for Class 9 Maths NCERT Solutions for Class 9 Science

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - Key Features

Comprehensive and Step-by-Step: These maths chapter 4 class 9 solutions provide comprehensive coverage of the topic, presented in a step-by-step format for easy understanding.

Practice-Driven: They offer a variety of exercises, including illustrative examples and practical applications, along with additional practice questions to strengthen problem-solving skills.

Accurate and Exam-Oriented: The ch 4 maths class 9 solutions are accurate, exam-oriented, and readily accessible, making them an ideal resource for self-study and preparation for class assessments and board exams.

## NCERT Books and NCERT Syllabus

Keep Working Hard & Happy Learning!

1. What are the important topics in chapter Linear Equations in Two Variables ?
• Linear equations, solutions of linear equations
• The graph of linear equation in two variables,
• Equations of lines parallel to x-axis and y-axis are the important topics of this chapter.
2. Is it necessary to learn all the questions in linear equations in two variables questions?

Yes, practicing all the questions and formulas related to maths class 9 chapter 4  is essential to perform well in CBSE exams. Careers360 website provides accurate and easy-to-understand solutions, which can be beneficial for students to score higher marks. Apart from exam preparation, these solutions can also assist in solving homework and assignments. Therefore, students can start practicing NCERT Solutions for Class 9 Maths Chapter 4 to improve their overall performance.

3. Where can I find the complete class 9 maths chapter 4 solutions?

Here you will get the detailed NCERT solutions for class 9 maths  by clicking on the link. you can practice linear equations class 9 NCERT solutions which provide you indepth understanding of concepts that ultimately lead to score well in the exam. Also for ease you can study linear equations in two variables class 9 pdf both online and offline mode.

4. How does the NCERT solutions are helpful ?

NCERT solutions are provided in a very detailed manner which will give the conceptual clarity to the students. Also, they can take help from these solutions if they are not able to solve on their own

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